.
A tank containing liquid is filled with 40L of 70% salt solution. What volume of the solution to be taken
and be filled up with 10L of water to make the concentration 50% salt solution?
~~~~~~~~~~~~~~~~
The maximum possible concentration of salt in water is about 28%.
Then saturation occurs, and no more salt dissolves in water.
See this Wikipedia article https://en.wikipedia.org/wiki/Saline_water
If to forget (to neglect) this saturation effect, then the problem can be solved in this way :
Let W be the volume (in liters) of the 70% salt solution to take it off.
After taking off W liters of the 70% solution, you have (40-W) liters of the 70% solution, and you add 10 liters of water.
So, you have THIS "concentration" equation
= 0.5.
It is your basic equation for the problem.
Simplify and solve for W.
0.7*(40-W) = 0.5*(50-W)
0.7*40 - 0.7W = 0.5*50 - 0.5W
0.7*40 - 0.5*50 = 0.7W - 0.5W
28 - 25 = 0.2W
3 = 0.2W
W = = 15.
ANSWER. 15 liters of the 70% solution should be taken off and then 10 liters of water added.
Solved.
--------------------
There is a bunch of introductory lessons in this site, covering various types of mixture problems
- Mixture problems
- More Mixture problems
- Solving typical word problems on mixtures for solutions
- Word problems on mixtures for antifreeze solutions (*)
- Word problems on mixtures for alloys
- Typical word problems on mixtures from the archive
Read them and become an expert in solution the mixture word problems.
Your problem is very similar to relevant problems on antifreeze mixtures.
See the lesson marked (*) in the list above.
Also, you have this free of charge online textbook in ALGEBRA-I in this site
- ALGEBRA-I - YOUR ONLINE TEXTBOOK.
The referred lessons are the part of this online textbook under the topic "Mixture problems".
Save the link to this online textbook together with its description
Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson
to your archive and use it when it is needed.